The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is

$36 x^{2} + 9 y^{2} = 4 l^{2}$

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$36 x^{2} + 9 y^{2} = l^{2}$

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$9 x^{2} + 36 y^{2} = 4 l^{2}$

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3x+5y-7=0

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Solution

The correct option is C
$9 x^{2} + 36 y^{2} = 4 l^{2}$

According to the figure

$P (h, k) d i v i d e s A B i n t h e r a t i o 1 : 2 \Rightarrow (\frac{2 a}{3}, \frac{b}{3}) = (h, k) a = \frac{3 h}{2}, b = 3 k a^{2} + b^{2} = l^{2} \Rightarrow \frac{9 h^{2}}{4} + 9 k^{2} = l^{2} \Rightarrow 9 h^{2} + 36 k^{2} = 4 l^{2} ∴ l o c u s o f P I S 9 x^{2} + 36 y^{2} = 4 l^{2}$

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